Solving the Difference Equation $Y_{t+1} - 1.2Y_t = 0$

The given difference equation is:

$Y_{t+1} - 1.2Y_t = 0$

This is a first-order linear homogeneous difference equation. We can rewrite it as:

$Y_{t+1} = 1.2Y_t$

Finding the General Expression for $Y_t$

This recursive relationship implies that each term is $1.2$ times the previous term, forming a geometric progression.

Let's express $Y_t$ in terms of an earlier value, say $Y_1$:

• $Y_t = 1.2 Y_{t-1}$
• Substitute $Y_{t-1} = 1.2 Y_{t-2}$: $Y_t = 1.2 (1.2 Y_{t-2}) = (1.2)^2 Y_{t-2}$
• Continuing this pattern, we can generalize this to:

$Y_t = (1.2)^k Y_{t-k}$

To express $Y_t$ in terms of $Y_1$, we need the index of $Y$ to be $1$. We set $t-k = 1$, which means $k = t-1$. Substituting this value of $k$ into the general form:

$Y_t = (1.2)^{t-1} Y_{t-(t-1)}$

$Y_t = (1.2)^{t-1} Y_1$

This expression represents the general solution for $Y_t$ based on the value $Y_1$. Let's verify this with the options provided.

• Option 1: $Y_t = (1.2)^t Y_1$. If $t=1$, $Y_1 = 1.2 Y_1$, incorrect.
• Option 2: $Y_t = (1.2)^{t+1} Y_0$. If $t=0$, $Y_0 = 1.2 Y_0$, incorrect.
• Option 3: $Y_t = (1.2)^{t-1} Y_1$. If $t=1$, $Y_1 = (1.2)^0 Y_1 = Y_1$. If $t=2$, $Y_2 = (1.2)^1 Y_1 = 1.2 Y_1$, which matches $Y_{t+1} = 1.2 Y_t$ for $t=1$. This is correct.
• Option 4: $Y_t = (1.2)^{t-1} Y_0 + C$. This form includes a constant $C$, which is typical for non-homogeneous equations. The given equation is homogeneous. Incorrect.

Therefore, the correct general expression is $Y_t = (1.2)^{t-1} Y_1$.